A projectile is fired from ground with an initial velocity that has a vertical component of 20m/s and a horizontal component of 30m/s. Using g = 10m/s2, the distance from launching to landing points is:

Correct answer: D. 120m

  • A. 40m
  • B. 80m
  • C. 60m
  • D. 120m

Explanation

The projectile is launched with a vertical velocity of 20 m/s and a horizontal velocity of 30 m/s. To find the time of flight, we first calculate the time to reach maximum height using the formula t = v/g. Here, v is 20 m/s and g is 10 m/s2, which results in a time to peak of 2 seconds. Therefore, the total time of flight is twice this value, amounting to 4 seconds. Next, we calculate the range using the horizontal velocity: Range = horizontal velocity × total time of flight. This gives us Range = 30 m/s × 4 s = 120 m. Options A and B underestimate the range due to miscalculations regarding time of flight. Option C, while it appears plausible, inaccurately simplifies the relationship between time of flight and horizontal distance, leading to an incorrect conclusion. Therefore, the correct range is found in Option D at 120m.

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Displacement, velocity and acceleration describe motion, including the horizontal and vertical components of projectile motion. The chapter connects these quantities with Newton's laws, momentum and impulse, then applies conservation of momentum to collisions, distinguishing elastic collisions from inelastic ones.

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